# In a PID every nonzero prime ideal is maximal

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal

Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which every ideal $A$ is of the form $\langle a \rangle = \{ar~~ | ~~r \in R\}$

Let $\langle a \rangle$ be a prime ideal $\implies R/A$ is an integral domain.

A finite integral domain is a field . Hence, if we prove that $R/A$ is finite, then $R/A$ is a field $\implies A$ is a maximal ideal.

Now, $\langle a \rangle = \{ar~~ | ~~r \in R\}$

Since, R is an integral domain, there are no zero divisors and cancellation is allowed $\implies ar_1 = ar_2 \implies r_1=r_2 \implies ar_i$ maps to a different member of $R$ for each different $r_i \implies \langle a \rangle$ represents the elements of $R$ in some random order.

$\implies \langle a \rangle = R$ and hence, $R/A \approx {0}$ is finite and hence $A$ is maximal.

Is my attempt correct?

• No, this is impossible. Consider $\Bbb Q[x]$ which is a PID because it is a Euclidean domain (polynomials over a field). Irreducible polynomials generate the maximal ideals, but the quotient is not a finite field. Jul 12, 2014 at 18:01
• So you proved $\;\langle a\rangle=R\;$ and thus any prime ideal is the whole ring? This is obviously wrong...And besides this, $\;A=\langle a\rangle =R\implies R/A=R/R\cong\{0\}\;$ Jul 12, 2014 at 18:01
• @AdamHughes and Timbuc can you please pin point in the proof where a mistake could have been committed? Jul 12, 2014 at 18:03
• Your error is assuming that just because the map: $$\begin{cases}f_a: R\to R \\ r\mapsto ra\end{cases}$$ is bijective just because it is injective (domain implies injective). In fact, injective implies bijective if $R$ is finite, but not all integral domains are finite, take $\Bbb Z$, look at the multiplication by $2$ map, you get all the even numbers, but it clearly does not permute the elements bijectively. Jul 12, 2014 at 18:08

To prove this we use that $PID$ are UFDs. Then Let $\mathfrak{p}=(p)$ be a prime ideal of $R$. Assume there is an ideal $\mathfrak{p}\subseteq\mathfrak{m}=(m)\subseteq R$. Then we would have that $m|p$, but then by the defintion of a prime element of a UFD this means that $m=p$ or $m=u$, a unit. Hence $\mathfrak{m}=(p)$ or $R$, proving maximality.

Edit: Since the op hasn't seen UFDs yet, here's a quick way around that:

Becaues $(p)\subseteq (m)$ we have that there is a $b\in R$ so that $ap=bm$ for every $ap\in (p)$. In particular $p=b_0m$. But then as $p$ is prime, either $m$ or $b_0$ must be a unit. the first case implies $\mathfrak{m}=R$ the second implies $\mathfrak{m}=\mathfrak{p}$, hence $\mathfrak{p}$ is maximal.

• In fact you prove that every PID is a UFD exactly in this way. Jul 12, 2014 at 18:11
• Isn't this often used in the proof that PID's are UFD's? I mean, you can very easily adapt the proof so you aren't using anything UFD specific (prove the chain that if $r$ is non-zero/a non-unit, $r$ irreducible $\implies$ $(r)$ prime $\implies$ $(r)$ maximal $\implies$ $r$ irreducible) Jul 12, 2014 at 18:12
• @Crostul You'd also need to prove that PID's are FD's, and also prove that the property that the above property and being a FD is equilivant to it being a UFD (where we usually define it as saying that prime factorisations exist and are unique up to units) Jul 12, 2014 at 18:14
• @Andrew D: I totally agree with you, I was stressing just on what you said. Jul 12, 2014 at 18:17
• @Crostul Ah, I see. Jul 12, 2014 at 18:18

Hint  Note that for principal ideals: $$\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$$,  i.e. $$(a)\supset (b)\iff a\mid b,\,$$ hence

$$\qquad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\! (p)\:\! \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\[.1em] &\iff&\, p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\[.1em] &\iff&\, p\ \ \text{ is irreducible}\\[.1em] &\iff& \!(p)\:\! \text{ is prime, } \ \text{by PID} \Rightarrow\text{UFD,}\ \text{ so ireducible = prime } \end{eqnarray}$$

Remark  PIDs are the UFDs of dimension $$\le 1,\,$$ i.e. where all prime ideals $$\ne 0\,$$ are maximal.

• @VHP I recomomend that you give some consideration to this viewpoint, since it may prove conceptually useful later in your studies. Jul 12, 2014 at 18:29
• Surely, I will. I just completed the Factor rings chapter in Gallian. It hasn't yet introduced the Unique Factorisation Domain as of yet, but will try to relate what you said when I start reading about it. Jul 12, 2014 at 18:37
• @VHP You need only the direction $(\Leftarrow)$ which uses only the simple direction of the final equivalence, i.e. that primes are irreducible. So no knowledge of UFDs is needed for that. Jul 12, 2014 at 19:17
• Could you explain the first equivalence that $(p)$ is maximal is equivalent to $(p)$ has no proper container $(a)$? Why isn't there some other type of ideal (not principal) that contains $(p)$? Apr 7, 2017 at 3:23
• @Maggie In a PID every ideal is principal. Apr 7, 2017 at 3:40

Your approach can't work. The ring $$\mathbb{Q}[x]$$, where $$\mathbb{Q}$$ is the field of rational numbers, is a PID. However, for the prime ideal $$P=\langle x\rangle$$ the quotient is $$\mathbb{Q}[x]/P\cong\mathbb{Q}$$, which is infinite.

Suppose $$P$$ is a nonzero prime ideal and $$P=\langle p\rangle$$. Now, let $$I=\langle a\rangle$$ be an ideal such that $$\langle p\rangle\subseteq\langle a\rangle$$. In particular $$p=ab$$ so $$ab\in P$$. Therefore, by definition of prime ideal, either $$a\in P$$ or $$b\in P$$.

If $$a\in P$$, then $$\langle a\rangle\subseteq\langle p\rangle$$, whence $$I=P$$.

If $$b\in P$$, then $$b=pc$$, so $$p=ab=apc$$ from which $$1=ac$$ and $$a$$ is invertible, whence $$I=R$$.

• How do we know that $p=apc$ imples that $1=ac$? I mean, we don't know for sure that $p$ has a multiplicative inverse so we can not simply cancel it out. So why does it still allowed to cancel it? Any clarification? thank you in advance.
– FNH
Aug 2, 2015 at 0:12
• @MathsLover We are in an integral domain: $p=apc$ is the same as $p(1-ac)=0$ and, since $p\ne0$, we conclude $1-ac=0$. Aug 2, 2015 at 7:58
• Oh, I see. Thank you very much :)
– FNH
Aug 3, 2015 at 18:41
• @egreg I'm a bit stuck on the last line of your proof, with $b \in P$.... I understand that $c = a^{-1}$, but why does that mean $I = R$? Would really appreciate the clarification!
– user311260
Sep 16, 2017 at 23:47
• @Cici Since $I$ contains an invertible element, namely $a$, we conclude $I=R$. Sep 17, 2017 at 8:28

Let $$R$$ be a PID and $$p(\neq 0)\in R$$ be a prime element (which is irreducible also in $$R$$. Let $$p=ab$$ then $$p|a$$ or $$p|b$$. If $$p|a \Rightarrow a=pk$$ for some $$k\in R$$, $$p=pkb \Rightarrow p(1-kb)=0 \Rightarrow kb=1$$ since $$R$$ is domain. So $$p$$ is irreducible) then $$(p)$$ is a prime ideal. Hence $$(p)$$ is maximal (as $$p$$ is irreducible) in the set of all principally generated ideals of $$R$$ i.e., in $$R.$$ Hence $$(p)$$ is maximal ideal of $$R$$.